Question: $f(t)=(t-5)^2-9$ 1) What are the zeros of the function? Write the smaller $t$ first, and the larger $t$ second. $\text{smaller }t=$
Explanation: $\begin{aligned} (t-5)^2-9&=0 \\\\ (t-5)^2&=9 \\\\ \sqrt{(t-5)^2}&=\sqrt{9} \\\\ t-5&=\pm 3 \\\\ t&=\pm3+5 \\\\ t={2}&\text{ or }t={8} \end{aligned}$ $f(t)$ is given in vertex form: $f(t)=(t-{5})^2{-9}$ So the vertex of the parabola is at $({5},{-9})$. In conclusion, $\begin{aligned} \text{smaller }t&=2 \\\\ \text{larger }t&=8 \end{aligned}$ The vertex of the parabola is at $(5,-9)$